How long (distance and time) would it take a vehicle to accelerate from a stop to 203 mph with an average acceleration of 2.93ft/s²

t = 1.68 min

S = 2.82 miles

Explanation:

Given,

The initial velocity of the vehicle, u = 0

The final velocity of the vehicle, v = 203 mph

The average acceleration of the vehicle, a = 2.93 ft/s²

                                                                       = 7191.82 miles/h²

Using the first equation of motion

                                      v = u + at

                                       t = (v - u) / a

                                         = (203 - 0) / 7191.82

                                         = 0.028 h

                                         = 100.8 s

Thus, the time taken by the vehicle to reach the final velocity is, t = 100.8 s

Using the third equations of motion

                          S = ut + 1/2 at²

Substituting the values

                          S = 0.5 x 7191.82 x (0.028)²

                             = 2.82 miles

Hence, the distance traveled by the vehicle, S = 2.82 miles

the answer would be 32*f